这个问题类似于将列表切割成子列表列表,但在我的情况下,我希望包含每个前一个子列表的最后一个元素,作为下一个子列表中的第一个元素.并且必须考虑到最后一个元素总是至少有两个元素.
例如:
list_ = ['a','b','c','d','e','f','g','h']
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3号子列表的结果:
resultant_list = [['a','b','c'],['c','d','e'],['e','f','g'],['g','h']]
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wim*_*wim 14
您链接的答案中的列表理解很容易通过简单地缩短传递给范围的"step"参数来支持重叠块:
>>> list_ = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
>>> n = 3 # group size
>>> m = 1 # overlap size
>>> [list_[i:i+n] for i in range(0, len(list_), n-m)]
[['a', 'b', 'c'], ['c', 'd', 'e'], ['e', 'f', 'g'], ['g', 'h']]
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此问题的其他访问者可能无法使用输入列表(可切片,已知长度,有限).这是一个基于生成器的解决方案,可以处理任意迭代:
from collections import deque
def chunks(iterable, chunk_size=3, overlap=0):
# we'll use a deque to hold the values because it automatically
# discards any extraneous elements if it grows too large
if chunk_size < 1:
raise Exception("chunk size too small")
if overlap >= chunk_size:
raise Exception("overlap too large")
queue = deque(maxlen=chunk_size)
it = iter(iterable)
i = 0
try:
# start by filling the queue with the first group
for i in range(chunk_size):
queue.append(next(it))
while True:
yield tuple(queue)
# after yielding a chunk, get enough elements for the next chunk
for i in range(chunk_size - overlap):
queue.append(next(it))
except StopIteration:
# if the iterator is exhausted, yield any remaining elements
i += overlap
if i > 0:
yield tuple(queue)[-i:]
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注意:此实现是从中复制的wimpy.util.chunks
.如果您不介意添加依赖项,则可以pip install wimpy
使用from wimpy import chunks
而不是复制粘贴代码.
more_itertools
有一个用于重叠迭代的窗口工具。
给定的
import more_itertools as mit
iterable = list("abcdefgh")
iterable
# ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
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代码
windows = list(mit.windowed(iterable, n=3, step=2))
windows
# [('a', 'b', 'c'), ('c', 'd', 'e'), ('e', 'f', 'g'), ('g', 'h', None)]
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如果需要,您可以None
通过过滤窗口来删除填充值:
[list(filter(None, w)) for w in windows]
# [['a', 'b', 'c'], ['c', 'd', 'e'], ['e', 'f', 'g'], ['g', 'h']]
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有关详细信息,另请参阅 more_itertools
文档more_itertools.windowed